Question

If $f\left(x\right)=1+nx+\frac{n(n-1)}{2}{x}^2+\frac{n(n-1)(n-2)}{6}{x}^3+\cdots +x^n$, where $n\in N,$ then $f^{\prime \prime }\left(1\right)$ is equal to

• A. $n(n-1{)}^{2n}$
• B. $n(n-1)2^{n-1}$
• C. $(n-1)2^{n-1}$
• D. $n(n-1)2^{n-2}$

Answer 1

The correct option is D $n(n-1)2^{n-2}$

$f\left(x\right)=(1+x{)}^n$
$f^{\prime }\left(x\right)=n(1+x{)}^{n-1}$
$f^{\prime \prime }\left(x\right)=n(n-1)(1+x{)}^{n-2}$
$f^{\prime \prime }\left(1\right)=n(n-1)\cdot 2^{n-2}$